I would probably say worse,

@surfer

lol I just looked at Political Rules. No wonder I missed it :)

Thanks

The global game has a ton of playing pieces, but there’s always room for improvement. Like many people, we 3d printed air and naval bases. I’m curious what pieces others have cannibalized from other A&A editions. This is what we use:

Major Industrial Complex - White complexes from Classic Minor Industrial Complex - Grey complexes from Revised China Artillery - UK artillery from Revised (same color) China Flying Tigers - UK fighter from Revised

@nishav

Yep.

Also a rocket from Cape Town can hit Cairo, 9000+ km away.

@Panther Thank you. An exact answer as always.

@SuperbattleshipYamato @Galendae @FranceNeedsMorePower
Here is an idea for a German Strategy

https://youtu.be/76tH7nevVW0?si=JgcqXg3ntXp-wpr2

@Graunie:

The USS Lexington took two torpedo hits and 3 bombs during the Battle of the Coral Sea and continued flight operations until the battle was over.  Some explosions later sunk it, but that’s pretty impressive.  That was before they started putting torpedo hulls on the carriers.

The USS Yorktown dodged 8 torpedos in that battle.  Got all busted up with bombs and scrambled back for the defense of Midway after a few days worth of repairs.

If they were so easy to kill why didn’t they just kill them before they did so much damage to the Japanese fleet?  When the reality is every ship and plane is pretty much focused on taking out said carriers.

They are two hit pieces for sure.

Amen.

I also like Gargantua’s idea of having both carrier types available to purchase.  You could consider one an escourt carrier and the other the main battle carrier.  I am guessing Imperious Leader can tell us what their abreviations should be (CV, CVE, ABCD, etc.).

If you only have one type of troop, it’s a binomial distribution:

Prob(k) = (n choose k) * p^k * (1-p)^(n-k),

where k is the number of hits, n is the total number of units, and p is the probability of a hit.

If you have different units, you need a product of binomial distribution for each permutation of possible events.

So, for the example of two tanks and two infantry, what is the probability of three hits? I’ll in words, then write it out:  You can have either 2 infantry hit, and one tank, or two tanks and one infantry.  Those are the only two ways that you can get two hits:

Prob(3) = (2 choose 2) * (1/6)^(2) * (5/6)^(2-2) * (2 choose 1) * (3/6)^(1) * (3/6)^(2-1) + (2 choose 1) * (1/6)^(1) * (5/6)^(2-1) * (2 choose 2) * (3/6)^(2) * (3/6)^(2-2)

Prob(3) = 1 * 1/36 * 1 * 2 * 3/6 * 3/6 + 2 * 1/6 * 5/6 * 1 * 9/36 * 1

Prob(3) = 18/1296 + 90/1298

Prob(3) = 108/1298

Now, if you have 3 INF, 2 TANKS, and 1 FGT, what is the probability of 2 hits?

Well, how many ways are there to make two hits?  I’ll write them out:

2 INF, 0 Tanks, 0 Fighters
1 I, 1 T, 0 F
1 I, 0 T, 1 F
0 I, 1 T, 1 F
0 I, 2 T, 0 F

An example of how one of these terms expands:

1 I, 1 T, 0 F:

(2 choose 1) * (1/6)^(1) * (5/6)^(2-1) * (2 choose 1) * (3/6)^(1) * (3/6)^(2-1) * (1 choose 0) * (3/6)^0 * (3/6)^(2-2)

If you want Prob(2), then you need to take the sum of the probabilities of all those possible ways to make 2 hits.

Clear as mud?